Answer:
Step-by-step explanation:
7. 1/4 of 84 is 21, then 1/3 of that is 7
Answer:
Step-by-step explanation:
Set 1
first of all we have to look at the negative numbers:
-99% = -0,99
-√80 = is almost 9
so :
-9,(9) ; -√80 ; -99%
then the positive numbers:
19% = 0,19
√9 = 3
1/9 = 0,11...
so
1/9 ; 19% ; √9
final answer:
-9,(9) ; -√80 ; -99% ; 1/9 ; 19%, √9
Set 2
descending order = from greatest to least
In this case we have to look at first at positive numbers
2/3 = 0,6
√23 = 4,...
20/3 = 6,...
23% = 0,23
so
20/3 ; √23 ; 2/3 ; 23%
negative numbers
-√34 = -5,...
so :
-4,(3) ; -√34
final answer:
20/3 ; √23 ; 2/3 ; 23% ; -4,(3) ; -√34
This is a right triangle. To find the length of the hypotenuse a^2 + b^2 = c^2.
21^2 + 18^2 = c^2
441 + 324 = c^2
765 = c^2
27.66 km


Cancel out the denominators.
(x+5)² + 18 = -30
(x+5)² = -30 - 18
(x+5)² = -48
The equation is impossible, we have a negative number.
There isn't a number that squared gives a negative number (the square is always positive).
√(x+5)² = +/- √-48
x+5 = +/- √-48
IMPOSSIBLE
You can write the result in imaginary numbers
x+5=+/- 4i√3
Answer:
- False
- True
-- False
-- True
-- True
Step-by-step explanation:
The points are
,
,
,
and
---- missing from the question
Given

Required
Determine if each of the points would be on 
To do this, we simply substitute the value of x and of each point in
.
(a)
In this case;
and 
becomes




<em>The point </em>
<em> won't be on the graph because the corresponding value of y for </em>
<em> is </em>
<em></em>
(b) 
In this case;


becomes





<em>The point </em>
<em> would be on the graph because the corresponding value of y for </em>
is 
(c) 
In this case:

becomes





<em>The point </em>
<em> wouldn't be on the graph because the corresponding value of y for </em>
<em> is </em>
<em></em>
(d) 
In this case;

becomes


<em>The point </em>
<em> would be on the graph because the corresponding value of y for </em>
is 
(e)
In this case:
; 
becomes




<em>The point </em>
<em> would be on the graph because the corresponding value of y for </em>
is 